Problem: Multiply the following rational expressions and simplify the result. $\dfrac{z^8+11z^4+30}{18z^2+12z+2} \cdot \dfrac{24z^2+8z}{z^2+6z}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $z^8+11z^4+30$, of the first expression can be factored as $(z^4+5)(z^4+6)$ using the sum-product pattern. The denominator, $18z^2+12z+2$, of the first expression can be factored as $2(3z+1)(3z+1)$ by factoring out $2$ and using the perfect square pattern. The numerator, $24z^2+8z$, of the second expression can be factored as $8z(3z+1)$ by factoring out $8z$. The denominator, $z^2+6z$, of the second expression can be factored as $z(z+6)$ by factoring out $z$. Now the product looks as follows: $\dfrac{(z^4+5)(z^4+6)}{2(3z+1)(3z+1)}\cdot\dfrac{8z(3z+1)}{z(z+6)}$ To find the product of two rational expressions, we multiply across, and simplify. [What's that?] $\begin{aligned} &\phantom{=} \dfrac{(z^4+5)(z^4+6)}{2(3z+1)(3z+1)}\cdot\dfrac{8z(3z+1)}{z(z+6)} \\\\\\ &= \dfrac{(z^4+5)(z^4+6) \cdot 8z(3z+1)}{2(3z+1)(3z+1) \cdot z(z+6)} &\text{Multiply across.}\\\\\\ &=\dfrac{(z^4+5)(z^4+6) \cdot 4\cdot {\cancel{2}} {\cancel{z}}{\cancel{(3z+1)}}}{{\cancel{2}}{\cancel{(3z+1)}}(3z+1) \cdot {\cancel{z}}(z+6)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{4(z^4+5)(z^4+6)}{(3z+1)(z+6)}\end{aligned}$ Therefore, the simplified form of the product is $\dfrac{4\left(z^4+6\right)\left(z^4+5\right)}{\left(3z+1\right)\left(z+6\right)}$, which is equivalent to $\dfrac{4z^8+44z^4+120}{3z^2+19z+6}$.